How do I use the exponential shift to solve the linear differential equation y'' + 7y' +12y = 2(e^-3)(cos2x)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

How do I use the exponential shift to solve the linear differential equation y'' + 7y' +12y = 2(e^-3)(cos2x)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

im looking to find the general solution y = yc +yp. so far i have solved the homogenous part of the equation yc. yc = Ae^-3x + Be^-4x. i cant solve for yp yet.
ok
Let me do it on paper first. I think it would be murder typing it out.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

haha thank you! any help will go a long way
i'll continue to try draw something out of my mess of numbers and letters while i wait
i also figured that the derivative operator 'D' helps so that the LHS = (D^2 + 7D +12)y
You do mean e^{-3x}, don't you?
yes! sorry typo
When I get to factors that need expanding, I'm going to get Mathematica to do that (save time).
ok no worries. if i dont follow il just let u know
So what happened, your lecturer just dump theory on you with no demonstration?
he demonstated with the d.e. y'' + 6y' + 13y = x(e^3x)(sin2x) however that takes y = imaginary(Y) and also takes into account the complex numbers. As my roots are reall -3, -4 and i take y=real(Y) instead, i dont know what yp to try
he pretty much chalk vomits all over the board without going into the specifics of how he gets to places
Not trained to teach...
its really quite frustrating... luckily we are changing lecturers next week.. sorry for any hassle im causing you!
just give me a fan point and i'll be happy
wow thanks! im just gona go through it and check i understand.. haha ok sure how do i do that?
just click my 'become a fan' link...if there isn't one there, refresh the page...should work then
ps. is there a place to download mathmatica for free? ok right on it
www.wolframalpha.com The guy who made mathematica made this site. It can do a lot of things like integration, solving differential equations, plotting, etc.
fanned. thanks a million! will i still be able to ask help from you if i have any more troubles..
this is wolfram's solution to your equation http://www.wolframalpha.com/input/?i=y%27%27-2y%27%2B5y+%3D+16x^3*e^%283x%29+++
woops...no it's not...wait.
haha! i was like crap.. i did notttttt get that
http://www.wolframalpha.com/input/?i=y%27%27%2B7y%27%2B12y+%3D+2e^%28-3%29cos%282x%29
haha ur awesome at pen to paper but im afraid you did the same typo as me last time. theres an 'x' in e^-3x. however the graphs dont look how i thought they would..
graphs in the correct version*
ah...typo...
yep :) i thought it would have more of a decaying oscillation as it is overdamped.. is it because of the negative exponential? why?
So, am I right in understanding you've finished deriving the solution?
And you're reading off that and not Wolfram?
sorry no. i was comparing my notes to the graphs of wolfram.. but! i havent read through the wolfram content yet.. just making observations. if you would like you can continue helping other students.. and im guessing when i need help again it will notify you?
Yes, I get an e-mail if you respond to a thread I've been in. Are you fine to derive the rest?
If an error's been made on my part, it's because it's after 1am where I am. If you need help in the next ~12-24hrs, I may not be around. You'd should try your luck with someone else.
i should be ok for now thanks :) it takes me quite a while longer than you to come up with a solution
yep me too. 1:22am. ok do u recommend anyone else?
I don't know the particular skill level of people on here. You may have to post and cross fingers.
Is this an assignment due tomorrow or something?
ahah yeh you caught me out. due 2pm. i wish i could say im a more dedicated student working til the early hours of the morning doing maths... ah the thrills of life!
Well, since your timezone's mine, if I see something from you in the morning and I can respond, I will. My day will be hell busy, though.
Anyways, good luck. I seriously have to hit the sack.
thanks you are a hero! night!

Not the answer you are looking for?

Search for more explanations.

Ask your own question